Question

How do you evaluate the definite integral by the limit definition given `int 6dx` from [4,10]?

Answer 1

`36`

Explanation:

Geometric Approach

This is the equivalent of drawing a rectangle whose boundaries are the line `y = 6`, `y = 0`, `x = 4` and `x = 10`.

This rectangle would therefore have a length of `6` and a height of `6`, which would have an area of `36` square units.

Since the definite integral is in fact a measure of area, this is the answer.

Using the limit definition

The limit definition of a definite integral is `int_a^bf(x)dx = lim_(n->oo) sum_(i = 1)^n f(c_i)Delta x_i`.

The parameter `Delta x_i` is given by `(b - a)/n`. So, here we would have `(10 - 4)/n = 6/n`.

`c_i` is given by `a+iDeltax_i = 4+(6i)/n`.

Therefore:

`int_4^10 6dx = lim_(n-> oo)sum_(i = 1)^n f(c_i) Delta x_i`

`=lim_(n->oo) sum_(i = 1)^n f (4+(6i)/n) xx 6/n`

Since `f(4+(6i)/n) = 6`:

`=lim_(n->oo) sum_(i = 1)^n 6 xx 6/n`

`=lim_(n->oo) sum_(i = 1)^n 36/n`

Use the formula `sum_(i = 1)^n = n`.

`=lim_(n-> oo) n(36)/n`

`=lim_(n->oo) 36`

`= 36`

Same answer!

Calculus Approach

We integrate using the formula `intx^ndx= (x^(n + 1))/(n + 1) + C`.

`int(6)dx= 6(x^(0 + 1))/(0 + 1) = 6x`

We now rewrite in proper notation, and evaluate using the second fundamental theorem of calculus, which is that `int_a^b(F(x)) = f(b) - f(a)`, where `f(x)` is the antiderivative of `F(x)`:

`[6x]_4^10 = 6(10) -6(4) = 60 - 24 = 36`

Same answer as before, just using a different method.

Hopefully this helps!